Integrand size = 36, antiderivative size = 111 \[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{k} x}{\sqrt [3]{(1-x) x (1-k x)}}}{\sqrt {3}}\right )}{\sqrt [3]{k}}+\frac {\log (x)}{2 \sqrt [3]{k}}+\frac {\log (1-(1+k) x)}{2 \sqrt [3]{k}}-\frac {3 \log \left (-\sqrt [3]{k} x+\sqrt [3]{(1-x) x (1-k x)}\right )}{2 \sqrt [3]{k}} \] Output:
1/2*ln(x)/k^(1/3)+1/2*ln(1-(1+k)*x)/k^(1/3)-3/2*ln(-k^(1/3)*x+((1-x)*x*(-k *x+1))^(1/3))/k^(1/3)+arctan(1/3*(1+2*k^(1/3)*x/((1-x)*x*(-k*x+1))^(1/3))* 3^(1/2))*3^(1/2)/k^(1/3)
Time = 15.18 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.15 \[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{k} x}{\sqrt [3]{k} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )-2 \log \left (-\sqrt [3]{k} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (k^{2/3} x^2+\sqrt [3]{k} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 \sqrt [3]{k}} \] Input:
Integrate[(2 - (1 + k)*x)/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - (1 + k)*x)),x]
Output:
(2*Sqrt[3]*ArcTan[(Sqrt[3]*k^(1/3)*x)/(k^(1/3)*x + 2*((-1 + x)*x*(-1 + k*x ))^(1/3))] - 2*Log[-(k^(1/3)*x) + ((-1 + x)*x*(-1 + k*x))^(1/3)] + Log[k^( 2/3)*x^2 + k^(1/3)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x ))^(2/3)])/(2*k^(1/3))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {2-(k+1) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(k+1) x)} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {2-(k+1) x}{\sqrt [3]{x} (1-(k+1) x) \sqrt [3]{k x^2-(k+1) x+1}}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (2-(k+1) x)}{(1-(k+1) x) \sqrt [3]{k x^2-(k+1) x+1}}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {\sqrt [3]{x}}{(1-(k+1) x) \sqrt [3]{k x^2-(k+1) x+1}}+\frac {\sqrt [3]{x}}{\sqrt [3]{k x^2-(k+1) x+1}}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (\int \frac {\sqrt [3]{x}}{((-k-1) x+1) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}+\frac {\sqrt [3]{1-x} x^{2/3} \sqrt [3]{1-k x} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {1}{3},\frac {5}{3},\frac {2 k x}{k+\sqrt {k^2-2 k+1}+1},\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}\right )}{2 \sqrt [3]{k x^2-(k+1) x+1}}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
Input:
Int[(2 - (1 + k)*x)/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - (1 + k)*x)),x]
Output:
$Aborted
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 1.24 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (k^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 k^{\frac {1}{3}} x}\right )-\frac {\ln \left (\frac {k^{\frac {2}{3}} x^{2}+k^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\ln \left (\frac {-k^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )}{k^{\frac {1}{3}}}\) | \(111\) |
Input:
int((2-(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x),x,method=_RETURNVERBO SE)
Output:
-1/k^(1/3)*(3^(1/2)*arctan(1/3*3^(1/2)*(k^(1/3)*x+2*((-1+x)*x*(k*x-1))^(1/ 3))/k^(1/3)/x)-1/2*ln((k^(2/3)*x^2+k^(1/3)*((-1+x)*x*(k*x-1))^(1/3)*x+((-1 +x)*x*(k*x-1))^(2/3))/x^2)+ln((-k^(1/3)*x+((-1+x)*x*(k*x-1))^(1/3))/x))
Timed out. \[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\text {Timed out} \] Input:
integrate((2-(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x),x, algorithm="f ricas")
Output:
Timed out
\[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\int \frac {k x + x - 2}{\sqrt [3]{x \left (x - 1\right ) \left (k x - 1\right )} \left (k x + x - 1\right )}\, dx \] Input:
integrate((2-(1+k)*x)/((1-x)*x*(-k*x+1))**(1/3)/(1-(1+k)*x),x)
Output:
Integral((k*x + x - 2)/((x*(x - 1)*(k*x - 1))**(1/3)*(k*x + x - 1)), x)
\[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\int { \frac {{\left (k + 1\right )} x - 2}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (k + 1\right )} x - 1\right )}} \,d x } \] Input:
integrate((2-(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x),x, algorithm="m axima")
Output:
integrate(((k + 1)*x - 2)/(((k*x - 1)*(x - 1)*x)^(1/3)*((k + 1)*x - 1)), x )
\[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\int { \frac {{\left (k + 1\right )} x - 2}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (k + 1\right )} x - 1\right )}} \,d x } \] Input:
integrate((2-(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x),x, algorithm="g iac")
Output:
integrate(((k + 1)*x - 2)/(((k*x - 1)*(x - 1)*x)^(1/3)*((k + 1)*x - 1)), x )
Timed out. \[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\int \frac {x\,\left (k+1\right )-2}{\left (x\,\left (k+1\right )-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}} \,d x \] Input:
int((x*(k + 1) - 2)/((x*(k + 1) - 1)*(x*(k*x - 1)*(x - 1))^(1/3)),x)
Output:
int((x*(k + 1) - 2)/((x*(k + 1) - 1)*(x*(k*x - 1)*(x - 1))^(1/3)), x)
\[ \int \frac {2-(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} (1-(1+k) x)} \, dx=\left (\int \frac {x}{x^{\frac {4}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}} k +x^{\frac {4}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}}-x^{\frac {1}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}}}d x \right ) k +\int \frac {x}{x^{\frac {4}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}} k +x^{\frac {4}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}}-x^{\frac {1}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}}}d x -2 \left (\int \frac {1}{x^{\frac {4}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}} k +x^{\frac {4}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}}-x^{\frac {1}{3}} \left (k \,x^{2}-k x -x +1\right )^{\frac {1}{3}}}d x \right ) \] Input:
int((2-(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x),x)
Output:
int(x/(x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)*k*x + x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)*x - x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)),x)*k + int(x/ (x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)*k*x + x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)*x - x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)),x) - 2*int(1/(x**(1 /3)*(k*x**2 - k*x - x + 1)**(1/3)*k*x + x**(1/3)*(k*x**2 - k*x - x + 1)**( 1/3)*x - x**(1/3)*(k*x**2 - k*x - x + 1)**(1/3)),x)